Question

Solve the initial-value problem y = 24 – 3yy(1) = 1.

Answer 1

Answer:

$y = 8 -7e^{-3(x-1)}$

Step-by-step explanation:

Assuming that the equation is $y '= 24 -3y$ with initial condition $y(1) = 1$. We have,

$\frac{dy}{dx} + 3y = 24$, hence we can say that $P (x) = 3$ and $Q (x) = 24$ in the general form of the first order linear differential equation:

$\frac{dy}{dx} + P(x)y = Q(x)$

The integrating factor is given by:

$e^{\int{3} \, dx } = e^{3x}$. Thus, multiplying the entire equation by the integrating factor:

$e^{3x}\frac{dy}{dx} + 3e^{3x}y = 24e^{3x}$. This means that:

$\frac{d[e^{3x}y]}{dx} = 24e^{3x}$

$e^{3x}y = 8e^{3x} + C$ then

$y = 8 + Ce^{-3x}$. Applying the initial condition:

$C = -7e^{3}$ and therefore, $y = 8 -7e^{-3(x-1)}$

Assuming that the equation is $y = 24 -3y'$ with initial condition $y(1) = 1$. We have,

$\frac{dy}{dx} + \frac{1}{3}y = 8$, hence we can say that $P (x) = \frac{1}{3}$ and $Q (x) = 8$ in the general form of the first order linear differential equation:

$\frac{dy}{dx} + P(x)y = Q(x)$

The integrating factor is given by:

$e^{\int{\frac{1}{3}} \, dx } = e^{\frac{x}{3}}$. Thus, multiplying the entire equation by the integrating factor:

$e^{\frac{x}{3}}\frac{dy}{dx} + \frac{1}{3}e^{\frac{x}{3}}y = 8e^{\frac{x}{3}}$. This means that:

$\frac{d[e^{\frac{x}{3}}y]}{dx} = 8e^{\frac{x}{3}}$

$e^{\frac{x}{3}}y = 24e^{\frac{x}{3}} + C$ then

$y = 24 + Ce^{-\frac{x}{3}}$. Applying the initial condition:

$C = -23e^{\frac{1}{3}}$ and therefore, $y = 24 -23e^{-\frac{1}{3}(x-1)}$